Problem: Let $g(x)=\ln(x^4)$. Find $g'(x)$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac1{\ln(x)}+4x^3$ (Choice B) B $\dfrac{1}{x^4}$ (Choice C) C $4\ln(x^3)$ (Choice D) D $\dfrac4x$
Solution: $g(x)$ is a composition of two, more basic, functions: $x^4$ and $\ln(x)$. In other words, suppose $u(x)=x^4$ and $v(x)=\ln(x)$, then $g(x)=v\Bigl(u(x)\Bigr)$, or $(v\circ u)(x)$. Therefore, the derivative of $g$ can be found using the chain rule : $\begin{aligned} \dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&=\dfrac{dv}{du}\cdot\dfrac{du}{dx} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x) \end{aligned}$ Finding $v'\Bigl(u(x)\Bigr)$ $v(x)=\ln(x)$, and therefore $v'(x)=\dfrac1x$. Now we plug $u(x)=x^4$ into $v'$ : $\begin{aligned} v'\Bigl(u(x)\Bigr)&=v'\Bigl(x^4\Bigr) \\\\ &={\dfrac1{x^4}} \end{aligned}$ Finding $u'(x)$ $u(x)=x^4$, and therefore $u'(x)={4x^3}$. Putting things together $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}\left(\ln(x^4)\right) \\\\ &=\dfrac{d}{dx}\left[v\Bigl(u(x)\Bigr)\right]&&\gray{\text{Let }u(x)=x^4\text{, }v(x)=\ln(x)} \\\\ &=v'\Bigl(u(x)\Bigr)\cdot u'(x)&&\gray{\text{The chain rule}} \\\\ &={\dfrac1{x^4}}\cdot{4x^3} \\\\ &=\dfrac4x \end{aligned}$ In conclusion, $g'(x)=\dfrac4x$.